Yes, it's strange. We can no longer edit our previous responses.
For me, it is not only "strange", but also,
"too bad". This means that (apart from any
'embarrassment' that incorrect posts will remain "permanently" on line) the data base, as a whole, as a
'resource' for someone who just
"surfs-in" (looking for information) will no longer be as
reliable as it could have been. This is unfortunate. Someone "surfing" the net may arrive at a post in some thread and think that what is written there is
correct without realizing that several posts later on a comment has been made explaining how that post was in fact
incorrect.
I was envisioning that this website would become a real reliable "source" of
accurate information. Now, I see that as far my own posting is concerned, this will only be possible with
additional 'care', over and above the usual amount, to make sure that posts are placed "correctly" at the onset (or shortly thereafter). Given my own
limits of "time" and "knowledge", such a constraint may prove to be too demanding.
_______________
Actually, I have no idea what you mean by this --
"suppose for definiteness – without loss of generality"?
Would you mind elaborating it?
Sometimes, in the midst of a mathematical proof, one reaches a stage where a certain proposition P(k) will hold for at least one value of k. This particular value of k, however, is
'unknown' but nevertheless
'definite'. (For example, in the case of your "solution" above, the proposition P(k) was simply "a ≠ a
k", and this
had to be true for at least one of k =1 or k =2.)
Moreover, it is sometimes the case that the
continuation of the proof proceeds in 'identical' fashion
regardless of the
particular value of k for which P(k) is true. (This was indeed the case for your "solution".) So, instead of saying that P(k) is true for some
'definite' value of k, say k = k
o, where k
o is
'unspecified', one says "suppose for
definiteness that P(1) is true", and since the proof is the
'same' for any
other 'choice' of k, one adds the remark "... without loss of generality".
The statement is, therefore, a sort of 'shorthand' which allows one to
bypass certain 'mechanical' details and go straight to the essential idea behind the proof.
_______________
Basically, what will the Hilbert space look if I build it with the eigenvectors of the Q operator? Since we know its eigenvectors are not square integrable in the real line, This hilbert space might be bigger than the space of square integrable functions. Some thorough knowledge of functional analysis and probability theory might be needed here.
In the "functional analysis" approach, one
begins with a Hilbert space of square-integrable functions
R →
C. The 'justification' for this comes about from the
Schrödinger equation (in "x-space") coupled with the
Born probability rule that ψ
*(x)ψ(x) is the "probability density", where the latter of these implies that the (physical) wavefunctions are all
square-integrable. Thus, the
probability P(I) of finding the particle in the (non-infinitesimal) interval I is given by
P(I) = (ψ, P
Iψ) ,
where P
I is the "projector" defined by
[P
Iψ](x) ≡
ψ(x) , x Є I
0 , otherwise
and we have defined an "inner product"
(φ, ψ) = ∫φ
*(x)ψ(x) dx .
This 'family' of projectors P
I already contains in it the idea of |q><q|, since they are connected by the simple relation
P
(a,b) = ∫
ab |q><q| dq .
... Now, let's look more closely at what you say:
This hilbert space might be bigger than the space of square integrable functions.
Here, you are suggesting the idea of "building" a space from the |q>'s in such a way that those objects themselves are
included in the space. I have never thought abut such a proposition in any detail. Nevertheless, the
original Hilbert space would then be seen as "embedded" in a larger
'extended' vector space which would include the |q>'s (and whatever else).
For the record, you may want to know the
'technical' definition of a "Hilbert space"
H:
(i)
H is a "vector space";
(ii)
H has an "inner product" ( , );
(iii)
H is "complete" in the "induced norm" ║ ║ ≡ √( , );
(iv)
H is "separable".
The last of these is usually
not included in the definition. I have put it in here, since the Hilbert spaces of QM are always "separable". You may want to 'Google' some these terms or check at
mathworld or
Wikipedia, or the like.
Note that such a notion of an "extended" space is used in what is called a "rigged" Hilbert space. I do not know much about such a construction and, in particular, I am unsure as to what its
'utility' is from a 'practical' point of view.
There is also the "Theory of Distributions" (or "Distribution Theory"), which deals with this idea of "generalized" functions (i.e. "distributions") in a formally rigorous way.
_______________
Actually, that also can lead to another question. That justification brings me a Hilbert space with probability decomposition but not necessarily Complex value coefficiened. I think the need of Complex value coefficient seems to be explanable by the scattering of electron diffraction or its interference.
So far, we have been viewing the situation from a "static" perspective. As soon as we admit "motion" into the picture, then
complex-valued coefficients come into play by way of
necessity.
Think of a (time-independent) Hamiltonian, and the Schrödinger equation
ih
bar ∂
t|ψ(t)> = H|ψ(t)> .
With |φ
n> a basis of eigenkets such that H|φ
n> = E
n|φ
n> , we then have general solutions of the form
|ψ(t)> = ∑
n exp{ -iE
nt / h
bar } c
n |φ
n> .
There is
no way 'around' this. The coefficients
must be complex-valued.
Your example of "diffraction" or "interference" appears (to me) to be a
special case of this
general fact. On the other hand, we know that such problems can be 'treated' by the formalism of "classical optics", in which case the use of complex-valued coefficients is merely a matter of
'convenience', and
not one of
'necessity' (so, I'm not so sure that this is in fact a 'good' example).
_______________
In order to define Q | \psi >, you have to have a background manifold, then you can say Q | \psi > = q | \psi >, where q is not a constant.
You mean: Q|ψ
q> = q|ψ
q>, where q is not a constant.
---> Then this will lead to f(q) is not a square integrable function.
Yes. ... And as I mentioned above, "Distribution Theory" handles this
'difficulty' in a perfectly rigorous way.
_______________
